A sample of `NaHCO_(3) + Na_(2)CO_(3)` required 20 ml of HCl using phenolphthalein as indicator and 35ml more required if methyl orange is used as indicator . Then molar ratio of `NaHCO_(3)` to `Na_(2)CO_(3)` is

Options

A. `1/2`
B. `2/3`
C. `3/4`
D. `1/3`

Answer

To solve the problem, we need to find the molar ratio of \( \text{NaHCO}_3 \) to \( \text{Na}_2\text{CO}_3 \) based on the volumes of HCl required when using different indicators.

### Step-by-Step Solution:

1. **Understanding the Reactions**:
- When using phenolphthalein as an indicator, the reaction is:
\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
- When using methyl orange as an indicator, the reaction is:
\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]

2. **Volume of HCl Used**:
- For phenolphthalein, the volume of HCl used is 20 mL.
- For methyl orange, the total volume of HCl used is \( 20 \, \text{mL} + 35 \, \text{mL} = 55 \, \text{mL} \).

3. **Calculating Moles of HCl**:
- Let the molarity of HCl be \( M \).
- Moles of HCl used with phenolphthalein:
\[ n_1 = M \times \frac{20 \times 10^{-3}}{1} = 20M \times 10^{-3} \]
- Moles of HCl used with methyl orange:
\[ n_2 = M \times \frac{55 \times 10^{-3}}{1} = 55M \times 10^{-3} \]

4. **Setting Up the Molar Relationships**:
- From the reaction with phenolphthalein, the moles of \( \text{Na}_2\text{CO}_3 \) will equal the moles of HCl used:
\[ n_2 = n_1 \implies n_2 = 20M \times 10^{-3} \quad \text{(for } \text{Na}_2\text{CO}_3\text{)} \]
- From the reaction with methyl orange, the moles of \( \text{NaHCO}_3 \) will equal the moles of HCl used:
\[ n_2 = 55M \times 10^{-3} \quad \text{(for } \text{NaHCO}_3\text{)} \]

5. **Finding the Ratio**:
- Let \( x \) be the moles of \( \text{Na}_2\text{CO}_3 \) and \( y \) be the moles of \( \text{NaHCO}_3 \).
- From the equations:
\[ x = 20M \times 10^{-3} \]
\[ y = 55M \times 10^{-3} \]
- The molar ratio of \( \text{NaHCO}_3 \) to \( \text{Na}_2\text{CO}_3 \) is:
\[ \frac{y}{x} = \frac{55M \times 10^{-3}}{20M \times 10^{-3}} = \frac{55}{20} = \frac{11}{4} \]

### Final Answer:
The molar ratio of \( \text{NaHCO}_3 \) to \( \text{Na}_2\text{CO}_3 \) is \( \frac{11}{4} \).

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